Answer:
- 
Step-by-step explanation:
Given
f(x) =
- 
Evaluate f(19) by substituting x = 19 into f(x)
f(19) =
- 
=
- 
=
- 4
=
-
= - 
√214= 14.6287388383278
Calculator maybe??
Answer:
it has decreased the most in-between day 11 and day 6 because the graph has a little bit of a steeper drop. So there is less snow those days.
Step-by-step explanation:
Answer:



Step-by-step explanation:
Given
Let
A = Event of being a universal donor.
So:


Solving (a): Mean and Standard deviation.
The mean is:



The standard deviation is:




Solving (b): P(x = 3)
The event is a binomial event an dthe probability is calculated as:

So, we have:



